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GNU's grep has the option --only-matching, which prints just the matching region of a regular expression. I'm on a Solaris 5.10 box without any GNU tools installed, and I'm trying to achieve the same thing. Example:

grep -Eo "[0-9]+ ms" *.log

Is there a sed or awk expression that can do the same thing?

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It would help if you showed what a line from the file looks like and what exactly you're trying to extract. – Dennis Williamson Feb 10 '10 at 23:32
up vote 2 down vote accepted
sed -n 's/.*\([0-9]\+ ms\).*/\1/p' *.log
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Thanks. Changing the + to * works. Solaris' sed doesn't support extended regular expressions. – brianegge Feb 11 '10 at 1:30
@brianegge: The equivalent to + when it isn't available is, in this case, [0-9][0-9]* – Dennis Williamson Sep 18 '13 at 15:35

Try using /usr/sfw/bin/ggrep

/usr/sfw/bin/ggrep -V grep (GNU grep) 2.5

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Nope. The sfw packages aren't installed, and I don't have access to install them. The only directory I can write to is /tmp, and it would be easier to copy the files back to a machine with gnu grep on it than to copy grep to /tmp. – brianegge Feb 11 '10 at 0:24

Solaris 5.10 includes bash 3.00, which has a modern regex engine. The follow command works, though it's quite a bit more verbose than a grep or sed solution.

cat *.log | while read line; 
  if [[ $line =~ "([0-9]+) ms" ]]; then 
    echo "${BASH_REMATCH[1]}"; 

(Formatted on multiple lines for readability)

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Bash regex handling changed for version 3.2. You should not quote the pattern because it will be treated as a literal string instead of a regex. Since unquoted patterns also work in Bash 3.0 and 3.1, you should use them there, too, for forward compatibility. Further, I recommend that the pattern be stored in a variable and the variable should be used on the right side of =~. For example: pattern='([0-9]+) ms'; if [[ $line =~ $pattern ]] – Dennis Williamson Sep 18 '13 at 15:33

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